Long Division Algorithm Proof












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As I was doing my homework today, a sudden thought popped into my head. Why does our long-division algorithm work and how can I prove it? Why does it the function the way it does? Why do we not do division starting from the right and going to the left?



Thanks










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$endgroup$

















    1












    $begingroup$


    As I was doing my homework today, a sudden thought popped into my head. Why does our long-division algorithm work and how can I prove it? Why does it the function the way it does? Why do we not do division starting from the right and going to the left?



    Thanks










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      As I was doing my homework today, a sudden thought popped into my head. Why does our long-division algorithm work and how can I prove it? Why does it the function the way it does? Why do we not do division starting from the right and going to the left?



      Thanks










      share|cite|improve this question









      $endgroup$




      As I was doing my homework today, a sudden thought popped into my head. Why does our long-division algorithm work and how can I prove it? Why does it the function the way it does? Why do we not do division starting from the right and going to the left?



      Thanks







      algebra-precalculus






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      asked Dec 18 '18 at 1:16









      Dude156Dude156

      572316




      572316






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          Because multiplication is distributive over addition.



          $(m+a) d = md + ad$



          So if $dx = k + m$ we can solve $x = frac {k+m}d = frac kd + frac md$



          So to figure out $frac Nd$ is to solve $dx = N$ and if we have $N= 10*M + N'$ than we have to solve $x = frac {10*M + N'}d = frac {10*M}d + frac {N'}d$ if it is easier.



          But, you ask, what about remainders?



          For and $M$ you have $M = q*d + r$ for some remainder.



          So if $N = 10M + N'$ and $M = q*d + r$



          $N = 10M + N' = 10(q*d + r) + N'$ so



          $frac Nd = 10q + frac {(10r + N')}d$



          And $10r + N'$ must *also have something in that form where $10r_1 + N' = vd + s$



          So $N = 10M + N'=$



          $10(q*d + r) + N'=$



          $10q*d + (10r + N')=$



          $10q*d + v*d + s$ so



          $frac Nd = frac {10q*d + v*d + s}d = frac {10q*d}d + frac {vd}d + frac sd = 10q + v + frac sd$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Solid Answer man, thanks!
            $endgroup$
            – Dude156
            Dec 18 '18 at 1:52



















          2












          $begingroup$

          So instead of dividing using long division lets instead just subtract off multiples until we get one that too small and count how many times we do it. So for example if I'm dividing $17$ by $3$ I would calculate $17-3=14$ and set my count to $1$. I would repeat this again and again with $14-3=11$ and setting the count to two. Counting in this fashion when the count is set to $5$ we are left with $2$ and low and behold $3cdot5 + 2 =17$ and we have division with remainder.



          But this becomes impracticable quickly. Maybe we're trying to divide $9843$ by $7$ and subtracting off each $7$ one at a time would be tedious and time consuming. So, instead we subtract them off in groups of $70$ and increase the counter by $10$ each time instead. In fact, we can do even better and subtract of $7000$ to start, and increase the counter by $1000$ then work on the smaller problem.



          So finally we ask this question - what's the best way to group the $7$s so that they do the most work and I have to do the least steps? Ideally we want to find the biggest multiple of $7$ that is smaller than $9843$, subtract that off, add the number of multiples to the counter, then continue with the smaller sub-problem, repeating at each stage.



          And now that we're subtracting in the most efficient way, digit by digit, we finally realize we're just doing the division algorithm as usual now. Given how tedious long division can be the method does reduce the work. You can structure your proof around this idea of subtracting and grouping so I'll leave you the pleasure of reconstructing the result for yourself.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Long division works, because you are subtracting multiples of the divisor.



            Square roots are a form of long division, for which the divisor changes. In this case, we suppose a^2 has already been subtracted, and the next divisor is 20a+_. When we put b, this is the difference between the squares of 10a+b and 10a.



            Modified forms of the algorithm allow one to handle large bases, like 120.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Decimal long division algorithm $ $ Write the dividend as $, n = a ,10^{large k}! +b ,$ where $, a> d = $ divisor. Next divide $,a,$ by $,d,$ yielding $,color{#c00}{a = q,d + r}. $ Then



              $$dfrac{n}d = dfrac{color{#c00}a,10^{large k} + b}d = dfrac{(color{#c00}{qd+r})10^{large k} + b}{d} = color{#c00}q, 10^{large k} +!!! underbrace{dfrac{color{#c00}r10^{large k}+b}d}_{largerm color{#0a0}{recurse} on this}qquad $$



              Next $rmcolor{#0a0}{recursively}$ apply the algorithm to the indicated fraction (with smaller numerator).



              Usually we choose $a$ minimal, but we can choose any value of $,a>d,$ (it may simplify division)






              share|cite|improve this answer











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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                Because multiplication is distributive over addition.



                $(m+a) d = md + ad$



                So if $dx = k + m$ we can solve $x = frac {k+m}d = frac kd + frac md$



                So to figure out $frac Nd$ is to solve $dx = N$ and if we have $N= 10*M + N'$ than we have to solve $x = frac {10*M + N'}d = frac {10*M}d + frac {N'}d$ if it is easier.



                But, you ask, what about remainders?



                For and $M$ you have $M = q*d + r$ for some remainder.



                So if $N = 10M + N'$ and $M = q*d + r$



                $N = 10M + N' = 10(q*d + r) + N'$ so



                $frac Nd = 10q + frac {(10r + N')}d$



                And $10r + N'$ must *also have something in that form where $10r_1 + N' = vd + s$



                So $N = 10M + N'=$



                $10(q*d + r) + N'=$



                $10q*d + (10r + N')=$



                $10q*d + v*d + s$ so



                $frac Nd = frac {10q*d + v*d + s}d = frac {10q*d}d + frac {vd}d + frac sd = 10q + v + frac sd$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Solid Answer man, thanks!
                  $endgroup$
                  – Dude156
                  Dec 18 '18 at 1:52
















                1












                $begingroup$

                Because multiplication is distributive over addition.



                $(m+a) d = md + ad$



                So if $dx = k + m$ we can solve $x = frac {k+m}d = frac kd + frac md$



                So to figure out $frac Nd$ is to solve $dx = N$ and if we have $N= 10*M + N'$ than we have to solve $x = frac {10*M + N'}d = frac {10*M}d + frac {N'}d$ if it is easier.



                But, you ask, what about remainders?



                For and $M$ you have $M = q*d + r$ for some remainder.



                So if $N = 10M + N'$ and $M = q*d + r$



                $N = 10M + N' = 10(q*d + r) + N'$ so



                $frac Nd = 10q + frac {(10r + N')}d$



                And $10r + N'$ must *also have something in that form where $10r_1 + N' = vd + s$



                So $N = 10M + N'=$



                $10(q*d + r) + N'=$



                $10q*d + (10r + N')=$



                $10q*d + v*d + s$ so



                $frac Nd = frac {10q*d + v*d + s}d = frac {10q*d}d + frac {vd}d + frac sd = 10q + v + frac sd$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Solid Answer man, thanks!
                  $endgroup$
                  – Dude156
                  Dec 18 '18 at 1:52














                1












                1








                1





                $begingroup$

                Because multiplication is distributive over addition.



                $(m+a) d = md + ad$



                So if $dx = k + m$ we can solve $x = frac {k+m}d = frac kd + frac md$



                So to figure out $frac Nd$ is to solve $dx = N$ and if we have $N= 10*M + N'$ than we have to solve $x = frac {10*M + N'}d = frac {10*M}d + frac {N'}d$ if it is easier.



                But, you ask, what about remainders?



                For and $M$ you have $M = q*d + r$ for some remainder.



                So if $N = 10M + N'$ and $M = q*d + r$



                $N = 10M + N' = 10(q*d + r) + N'$ so



                $frac Nd = 10q + frac {(10r + N')}d$



                And $10r + N'$ must *also have something in that form where $10r_1 + N' = vd + s$



                So $N = 10M + N'=$



                $10(q*d + r) + N'=$



                $10q*d + (10r + N')=$



                $10q*d + v*d + s$ so



                $frac Nd = frac {10q*d + v*d + s}d = frac {10q*d}d + frac {vd}d + frac sd = 10q + v + frac sd$.






                share|cite|improve this answer









                $endgroup$



                Because multiplication is distributive over addition.



                $(m+a) d = md + ad$



                So if $dx = k + m$ we can solve $x = frac {k+m}d = frac kd + frac md$



                So to figure out $frac Nd$ is to solve $dx = N$ and if we have $N= 10*M + N'$ than we have to solve $x = frac {10*M + N'}d = frac {10*M}d + frac {N'}d$ if it is easier.



                But, you ask, what about remainders?



                For and $M$ you have $M = q*d + r$ for some remainder.



                So if $N = 10M + N'$ and $M = q*d + r$



                $N = 10M + N' = 10(q*d + r) + N'$ so



                $frac Nd = 10q + frac {(10r + N')}d$



                And $10r + N'$ must *also have something in that form where $10r_1 + N' = vd + s$



                So $N = 10M + N'=$



                $10(q*d + r) + N'=$



                $10q*d + (10r + N')=$



                $10q*d + v*d + s$ so



                $frac Nd = frac {10q*d + v*d + s}d = frac {10q*d}d + frac {vd}d + frac sd = 10q + v + frac sd$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 1:33









                fleabloodfleablood

                73k22789




                73k22789












                • $begingroup$
                  Solid Answer man, thanks!
                  $endgroup$
                  – Dude156
                  Dec 18 '18 at 1:52


















                • $begingroup$
                  Solid Answer man, thanks!
                  $endgroup$
                  – Dude156
                  Dec 18 '18 at 1:52
















                $begingroup$
                Solid Answer man, thanks!
                $endgroup$
                – Dude156
                Dec 18 '18 at 1:52




                $begingroup$
                Solid Answer man, thanks!
                $endgroup$
                – Dude156
                Dec 18 '18 at 1:52











                2












                $begingroup$

                So instead of dividing using long division lets instead just subtract off multiples until we get one that too small and count how many times we do it. So for example if I'm dividing $17$ by $3$ I would calculate $17-3=14$ and set my count to $1$. I would repeat this again and again with $14-3=11$ and setting the count to two. Counting in this fashion when the count is set to $5$ we are left with $2$ and low and behold $3cdot5 + 2 =17$ and we have division with remainder.



                But this becomes impracticable quickly. Maybe we're trying to divide $9843$ by $7$ and subtracting off each $7$ one at a time would be tedious and time consuming. So, instead we subtract them off in groups of $70$ and increase the counter by $10$ each time instead. In fact, we can do even better and subtract of $7000$ to start, and increase the counter by $1000$ then work on the smaller problem.



                So finally we ask this question - what's the best way to group the $7$s so that they do the most work and I have to do the least steps? Ideally we want to find the biggest multiple of $7$ that is smaller than $9843$, subtract that off, add the number of multiples to the counter, then continue with the smaller sub-problem, repeating at each stage.



                And now that we're subtracting in the most efficient way, digit by digit, we finally realize we're just doing the division algorithm as usual now. Given how tedious long division can be the method does reduce the work. You can structure your proof around this idea of subtracting and grouping so I'll leave you the pleasure of reconstructing the result for yourself.






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  So instead of dividing using long division lets instead just subtract off multiples until we get one that too small and count how many times we do it. So for example if I'm dividing $17$ by $3$ I would calculate $17-3=14$ and set my count to $1$. I would repeat this again and again with $14-3=11$ and setting the count to two. Counting in this fashion when the count is set to $5$ we are left with $2$ and low and behold $3cdot5 + 2 =17$ and we have division with remainder.



                  But this becomes impracticable quickly. Maybe we're trying to divide $9843$ by $7$ and subtracting off each $7$ one at a time would be tedious and time consuming. So, instead we subtract them off in groups of $70$ and increase the counter by $10$ each time instead. In fact, we can do even better and subtract of $7000$ to start, and increase the counter by $1000$ then work on the smaller problem.



                  So finally we ask this question - what's the best way to group the $7$s so that they do the most work and I have to do the least steps? Ideally we want to find the biggest multiple of $7$ that is smaller than $9843$, subtract that off, add the number of multiples to the counter, then continue with the smaller sub-problem, repeating at each stage.



                  And now that we're subtracting in the most efficient way, digit by digit, we finally realize we're just doing the division algorithm as usual now. Given how tedious long division can be the method does reduce the work. You can structure your proof around this idea of subtracting and grouping so I'll leave you the pleasure of reconstructing the result for yourself.






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    So instead of dividing using long division lets instead just subtract off multiples until we get one that too small and count how many times we do it. So for example if I'm dividing $17$ by $3$ I would calculate $17-3=14$ and set my count to $1$. I would repeat this again and again with $14-3=11$ and setting the count to two. Counting in this fashion when the count is set to $5$ we are left with $2$ and low and behold $3cdot5 + 2 =17$ and we have division with remainder.



                    But this becomes impracticable quickly. Maybe we're trying to divide $9843$ by $7$ and subtracting off each $7$ one at a time would be tedious and time consuming. So, instead we subtract them off in groups of $70$ and increase the counter by $10$ each time instead. In fact, we can do even better and subtract of $7000$ to start, and increase the counter by $1000$ then work on the smaller problem.



                    So finally we ask this question - what's the best way to group the $7$s so that they do the most work and I have to do the least steps? Ideally we want to find the biggest multiple of $7$ that is smaller than $9843$, subtract that off, add the number of multiples to the counter, then continue with the smaller sub-problem, repeating at each stage.



                    And now that we're subtracting in the most efficient way, digit by digit, we finally realize we're just doing the division algorithm as usual now. Given how tedious long division can be the method does reduce the work. You can structure your proof around this idea of subtracting and grouping so I'll leave you the pleasure of reconstructing the result for yourself.






                    share|cite|improve this answer











                    $endgroup$



                    So instead of dividing using long division lets instead just subtract off multiples until we get one that too small and count how many times we do it. So for example if I'm dividing $17$ by $3$ I would calculate $17-3=14$ and set my count to $1$. I would repeat this again and again with $14-3=11$ and setting the count to two. Counting in this fashion when the count is set to $5$ we are left with $2$ and low and behold $3cdot5 + 2 =17$ and we have division with remainder.



                    But this becomes impracticable quickly. Maybe we're trying to divide $9843$ by $7$ and subtracting off each $7$ one at a time would be tedious and time consuming. So, instead we subtract them off in groups of $70$ and increase the counter by $10$ each time instead. In fact, we can do even better and subtract of $7000$ to start, and increase the counter by $1000$ then work on the smaller problem.



                    So finally we ask this question - what's the best way to group the $7$s so that they do the most work and I have to do the least steps? Ideally we want to find the biggest multiple of $7$ that is smaller than $9843$, subtract that off, add the number of multiples to the counter, then continue with the smaller sub-problem, repeating at each stage.



                    And now that we're subtracting in the most efficient way, digit by digit, we finally realize we're just doing the division algorithm as usual now. Given how tedious long division can be the method does reduce the work. You can structure your proof around this idea of subtracting and grouping so I'll leave you the pleasure of reconstructing the result for yourself.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 22 '18 at 6:04

























                    answered Dec 18 '18 at 1:49









                    CyclotomicFieldCyclotomicField

                    2,4431314




                    2,4431314























                        1












                        $begingroup$

                        Long division works, because you are subtracting multiples of the divisor.



                        Square roots are a form of long division, for which the divisor changes. In this case, we suppose a^2 has already been subtracted, and the next divisor is 20a+_. When we put b, this is the difference between the squares of 10a+b and 10a.



                        Modified forms of the algorithm allow one to handle large bases, like 120.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Long division works, because you are subtracting multiples of the divisor.



                          Square roots are a form of long division, for which the divisor changes. In this case, we suppose a^2 has already been subtracted, and the next divisor is 20a+_. When we put b, this is the difference between the squares of 10a+b and 10a.



                          Modified forms of the algorithm allow one to handle large bases, like 120.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Long division works, because you are subtracting multiples of the divisor.



                            Square roots are a form of long division, for which the divisor changes. In this case, we suppose a^2 has already been subtracted, and the next divisor is 20a+_. When we put b, this is the difference between the squares of 10a+b and 10a.



                            Modified forms of the algorithm allow one to handle large bases, like 120.






                            share|cite|improve this answer









                            $endgroup$



                            Long division works, because you are subtracting multiples of the divisor.



                            Square roots are a form of long division, for which the divisor changes. In this case, we suppose a^2 has already been subtracted, and the next divisor is 20a+_. When we put b, this is the difference between the squares of 10a+b and 10a.



                            Modified forms of the algorithm allow one to handle large bases, like 120.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 18 '18 at 1:37









                            wendy.kriegerwendy.krieger

                            5,84511427




                            5,84511427























                                0












                                $begingroup$

                                Decimal long division algorithm $ $ Write the dividend as $, n = a ,10^{large k}! +b ,$ where $, a> d = $ divisor. Next divide $,a,$ by $,d,$ yielding $,color{#c00}{a = q,d + r}. $ Then



                                $$dfrac{n}d = dfrac{color{#c00}a,10^{large k} + b}d = dfrac{(color{#c00}{qd+r})10^{large k} + b}{d} = color{#c00}q, 10^{large k} +!!! underbrace{dfrac{color{#c00}r10^{large k}+b}d}_{largerm color{#0a0}{recurse} on this}qquad $$



                                Next $rmcolor{#0a0}{recursively}$ apply the algorithm to the indicated fraction (with smaller numerator).



                                Usually we choose $a$ minimal, but we can choose any value of $,a>d,$ (it may simplify division)






                                share|cite|improve this answer











                                $endgroup$


















                                  0












                                  $begingroup$

                                  Decimal long division algorithm $ $ Write the dividend as $, n = a ,10^{large k}! +b ,$ where $, a> d = $ divisor. Next divide $,a,$ by $,d,$ yielding $,color{#c00}{a = q,d + r}. $ Then



                                  $$dfrac{n}d = dfrac{color{#c00}a,10^{large k} + b}d = dfrac{(color{#c00}{qd+r})10^{large k} + b}{d} = color{#c00}q, 10^{large k} +!!! underbrace{dfrac{color{#c00}r10^{large k}+b}d}_{largerm color{#0a0}{recurse} on this}qquad $$



                                  Next $rmcolor{#0a0}{recursively}$ apply the algorithm to the indicated fraction (with smaller numerator).



                                  Usually we choose $a$ minimal, but we can choose any value of $,a>d,$ (it may simplify division)






                                  share|cite|improve this answer











                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Decimal long division algorithm $ $ Write the dividend as $, n = a ,10^{large k}! +b ,$ where $, a> d = $ divisor. Next divide $,a,$ by $,d,$ yielding $,color{#c00}{a = q,d + r}. $ Then



                                    $$dfrac{n}d = dfrac{color{#c00}a,10^{large k} + b}d = dfrac{(color{#c00}{qd+r})10^{large k} + b}{d} = color{#c00}q, 10^{large k} +!!! underbrace{dfrac{color{#c00}r10^{large k}+b}d}_{largerm color{#0a0}{recurse} on this}qquad $$



                                    Next $rmcolor{#0a0}{recursively}$ apply the algorithm to the indicated fraction (with smaller numerator).



                                    Usually we choose $a$ minimal, but we can choose any value of $,a>d,$ (it may simplify division)






                                    share|cite|improve this answer











                                    $endgroup$



                                    Decimal long division algorithm $ $ Write the dividend as $, n = a ,10^{large k}! +b ,$ where $, a> d = $ divisor. Next divide $,a,$ by $,d,$ yielding $,color{#c00}{a = q,d + r}. $ Then



                                    $$dfrac{n}d = dfrac{color{#c00}a,10^{large k} + b}d = dfrac{(color{#c00}{qd+r})10^{large k} + b}{d} = color{#c00}q, 10^{large k} +!!! underbrace{dfrac{color{#c00}r10^{large k}+b}d}_{largerm color{#0a0}{recurse} on this}qquad $$



                                    Next $rmcolor{#0a0}{recursively}$ apply the algorithm to the indicated fraction (with smaller numerator).



                                    Usually we choose $a$ minimal, but we can choose any value of $,a>d,$ (it may simplify division)







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 18 '18 at 2:23

























                                    answered Dec 18 '18 at 2:11









                                    Bill DubuqueBill Dubuque

                                    213k29195654




                                    213k29195654






























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